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Free fall with resistance

  1. Oct 15, 2005 #1
    Here's the problem:
    Consider a particle of mass m whose motion starts from rest in a constant gravitational field. If a resisting force proportional to the square of the velocity (kmv2) is encountered, find the distance s the particle falls in accelerating from v0 to v1.

    I began with the equation m (dv/dt)=-mg-kmv2.
    From this, we see that dv = -g-kv2 dt.
    When I separate variables and integrate this however, I get a function that involves the inverse tangent. I would have to integrate a second time to find distance, correct?
    The book gives the distance s from v0 to v1 as 1/2k ln ((g-kv02)/(g-kv12)). What am I doing wrong???

    Any help greatly appreciated!!!
  2. jcsd
  3. Oct 15, 2005 #2
    Use the relation

    \ddot{x} = v \frac{dv}{dx}

    on the left hand side of your equation and you should find things work out :)
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